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Chapter 16: Problem 60

Two sources of sound are located on the x axis, and each emits power uniformly in all directions. There are no reflections. One source is positioned at the origin and the other at x 123 m. The source at the origin emits four times as much power as the other source. Where on the x axis are the two sounds equal in intensity? Note that there are two answers.

Short Answer

Expert verified
The sounds are equal in intensity at x = 82 m and potentially symmetric midpoint for broader structure context if applicable in design 123m back against configurations/symmetrical frames.

Step by step solution

01

Understanding Intensity and Power

The intensity of sound is defined as the power per unit area. For a point source, the intensity decreases with the square of the distance. If a source emits power \( P \), the intensity \( I \) at a distance \( r \) is given by \( I = \frac{P}{4\pi r^2} \). In this problem, since sound is emitted uniformly in all directions, the sound intensity depends on the distance from the sources.
02

Define Variables and Equations

Let's denote the power of the source at the origin as \( P_1 \) and the power of the source at \( x = 123 \) m as \( P_2 \), where \( P_1 = 4P_2 \). Let \( x \) be the position on the x-axis where intensities from both sources are equal. The intensity from the origin at any point \( x \) is \( I_1 = \frac{P_1}{4\pi x^2} \) and from \( x = 123 \) is \( I_2 = \frac{P_2}{4\pi (123-x)^2} \). We need \( I_1 = I_2 \).
03

Set Up the Equality of Intensities

Setting \( I_1 = I_2 \), we have: \[ \frac{P_1}{4\pi x^2} = \frac{P_2}{4\pi (123-x)^2} \] Simplify this equation by canceling the \( 4\pi \) terms: \[ \frac{P_1}{x^2} = \frac{P_2}{(123-x)^2} \] Since \( P_1 = 4P_2 \), substitute to get: \[ \frac{4P_2}{x^2} = \frac{P_2}{(123-x)^2} \].
04

Solve the Equation for x

Cancel \( P_2 \) from both sides: \[ \frac{4}{x^2} = \frac{1}{(123-x)^2} \] Cross-multiply to obtain: \[ 4(123-x)^2 = x^2 \] Simplify and solve for \( x \): \[ 4((123)^2 - 246x + x^2) = x^2 \] \[ 4 \times 15129 - 984x + 4x^2 = x^2 \] Simplifying: \[ 3x^2 - 984x + 60516 = 0 \].
05

Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 3 \), \( b = -984 \), and \( c = 60516 \). Calculate the discriminant \( b^2 - 4ac \): \[ (-984)^2 - 4 \times 3 \times 60516 = 968256 - 726192 = 242064 \]. Calculate \( x \): \[ x = \frac{984 \pm \sqrt{242064}}{6} \].
06

Find Solutions from Quadratic Formula

Compute \( \sqrt{242064} \approx 492 \). Substitute this alongside the plus/minus in the quadratic formula: \[ x = \frac{984 \pm 492}{6} \]. This gives two possible values for \( x \): \[ x = \frac{1476}{6} = 246 \] and \[ x = \frac{492}{6} = 82 \]. However, only the value \( x = 82 \) is within our range of concern on the x-axis. For symmetry and completeness, either check calculations or consider constraints to confirm points like \( -243 \) may be consistent but out of defined constraints (given symmetry).
07

Interpret Results

Both \( x = 82 \) m and symmetric counterpart can represent the points on the x-axis where the sound intensities from both sources are equal. Since our primary focus is within positive range values, and assumptions of symmetry need only physical bounds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power of Sound Sources
Sound sources can emit energy in the form of sound waves and the rate at which this energy is emitted is known as the power of the sound source. Sound power is distributed uniformly in all directions from the source, assuming there are no obstacles in the environment that may cause reflections.
This means the intensity of the sound, or how much power is passing through a unit area, depends on the distance from the sound source.
The unit of sound power is watts (W) and to calculate this, one can use the formula: \[ I = \frac{P}{A} \] where \( I \) is the intensity, \( P \) is the power, and \( A \) is the area through which the power is passing.
This is essential in understanding how sound behaves, especially in exercises involving sound sources placed at different locations.
Distance and Intensity Relationship
The intensity of a sound wave decreases as the distance from the source increases. This is because the energy spreads out over a larger area as you move away from the source, leading to a decrease in intensity. The relationship between distance and intensity is described by the inverse square law.
This law states that intensity is inversely proportional to the square of the distance from the source: \[ I \propto \frac{1}{r^2} \] Here, \(I\) is the intensity, and \(r\) is the distance from the source to the point of measurement.
Understanding this concept is crucial in solving problems related to sound intensity, where it's often required to find a point where two different sound intensities from different sources are equal.
Factors like power of the source and distance interplay to influence how intense a sound will be at any given point.
Quadratic Equations in Physics
In physics, quadratic equations often arise in problems involving distances and areas, such as in the sound intensity problem mentioned above. These equations are in the form: \[ ax^2 + bx + c = 0 \] These have solutions that can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.
Quadratic equations are particularly useful when determining distances in problems where variables are squared due to the relationship between intensity and distance (inverse square law).
By solving the quadratic equation, students can find the exact points of interest, such as where two sound sources have equal intensity on a common axis or path.
Mastery of solving quadratic equations aids deeply in tackling a broad range of physics and engineering challenges.

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Most popular questions from this chapter

An observer stands 25 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of the bullets is \(840 \mathrm{m} / \mathrm{s},\) and the air temperature is \(20^{\circ} \mathrm{C}\) . How far does each bullet travel before the observer hears the report of the rifle? Assume that the bullets encounter no obstacles during this interval, and ignore both air resistance and the vertical component of the bullets鈥� motion.

An amplified guitar has a sound intensity level that is 14 dB greater than the same unamplified sound. What is the ratio of the amplified intensity to the unamplified intensity?

To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.055-kg ball from the end of a wire. The wire has a length of 0.95 \(\mathrm{m}\) and a linear density of \(1.2 \times 10^{-4} \mathrm{kg} / \mathrm{m}\) . Using electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of 0.016 \(\mathrm{s}\) . The mass of the wire is negligible compared to the mass of the ball. Determine the acceleration due to gravity.

(a) A uniform rope of mass m and length L is hanging straight down from the ceiling. A small-amplitude transverse wave is sent up the rope from the bottom end. Derive an expression that gives the speed v of the wave on the rope in terms of the distance y above the bottom end of the rope and the magnitude g of the acceleration due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of 0.50 m and 2.0 m above the bottom end of the rope.

At a distance of 3.8 \(\mathrm{m}\) from a siren, the sound intensity is \(3.6 \times 10^{-2} \mathrm{W} / \mathrm{m}^{2}\) . Assuming that the siren radiates sound uniformly in all directions, find the total power radiated.
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